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Soft Skills

Functions and Graphs

In the new pattern of all B-school entrance exams, different exam boards giving an importance in this area. An average of 4 to 5 questions in almost all the top level exams (CAT,XAT etc)can expect from this chapter. XAT contains a large number of questions related to the user defined functions than the pure mathematical functions. A clear theoretical awareness of the concept of functions is required to solve most of the questions.

Here we are going through the following areas in Module 1.

Basic concepts of sets and relations

Definition of functions

Domain, co domain and range of a function

Understanding functions

Value of a function

Operations in functions

Classifying functions

Graphical representation

Maxima and minima in quadratic function

Basic concepts and Terminologies

Cross product of set A and B means set of all different possible mappings from set A to B. Set of all possible mappings can arrange as set of ordered pairs and this set of all relations can be expressed as A*B(read as A cross B)

A*B={(a,1), (a,2) , (b,1) , (b,2) , (c,1) , (c,2)}

Number of elements in A*B is the product of number of elements from set A and number of elements from set B.

A*B is also called the Cartesian product of the sets A and B.

Definition:

If A and B are two non-empty sets, then the Cartesian product of A and B is,

A*B = {(x , y) : x ε A and y ε B }

Either A or B is an empty set the A*B is also an empty set.

If (x, y) = (a, b) then x=a and y = b

Relation

A relation from set A to set B is the subset of Cartesian product A*B. Each of the selected ordered pairs from A * B describes a relation between the first element and the second element.

Domain, Co domain and Range of a relation

In a relation defined from set A to set B, the set of all first elements of the ordered pairs is called the domain of the relation and the set of all second elements (images) in the ordered pair is called the range of the relation. Set B is called the co-domain of the relation

i.e. Range ⊆ co-domain

Example

Let A= {1, 2, 3, 4} B= {1, 2, 3, 4, 5} and R is a relation defined from set A to set B

R= {(x, y): y = x+2, x ε A and y ε B}

Then R= {(1, 3), (2, 4), (3, 5)}

Domain of R = {1, 2, 3}

Range of R = {3, 4, 5}

Co-domain of R = {1, 2, 3, 4, 5}

Functions:

Functions are basically relations between two sets A and B.

When considering a relation from set A to set B, if each element in set A has one and only one image in set B, then the relation is called a function.

Let f is a function from set A to B then f(a) = b , where a ε A and b ε B and b is called image of a.

Illustrated examples:

A= {1, 2, 3, 4} and B= {a, b, c, d}

f1 = {(1,a),(2,a),(3,a),(4,a)} is a function. i.e. each element in set A has one and only one image in set B.

Domain of f1 = {1, 2, 3, 4}

Range of f1 = {a}

f2={(1,a),(2,b),(3,c),(4,d)} is a function.

Domain of f2 = {1, 2, 3, 4}

Range of f2 = {a, b, c, d}

f3 = {(1, a), (2, b), (3, c} is not a function because the element 1 and 2 from set A does not have any image in set B.

f4 = {(1, a),(1,b), (2, b),(2,b),(3,a), (4, d} is not a function because the element 4 in set A have more than one image in set B.

Value of a function:

Value of a function at certain input means the value of f(x) at certain value of x.

e.g. Let f(x) = 2x2 + 1

Then f (1) = 2(1)2 + 1 = 3

f (2) = 2(2)2 + 1 = 5

f (n+1) = 2(n+1)2 + 1

= 2(n2+2n+1) + 1

= 2n2+4n+2+1

= 2n2+4n+3

Operations on functions:

Functions are following the basic mathematical operations such as addition, subtraction, multiplication, division etc.

Let f(x) and g(x) are two different functions in x

(f + g)(x) = f(x) + g(x)

(f - g)(x) = f(x) - g(x)

(f * g)(x) = f(x) * g(x)

(f/g)(x) = f(x)/g(x) where g(x) ≠ 0

Illustrated examples

If f(x) = 2x and g(x) = x + 1 then

(f + g)(x) = f (x) + g (x)

= 2x + x + 1

= 3x + 1

(f-g)(x) = f(x) - g(x)

= 2x - (x+1)

= x-1

(f * g)(x) = f(x) * g(x)

= 2x(x+1)

= 2x2+2x

(f/g)(x) = f(x)/g(X) = 2x/x+1

Real Function:

Let f be a function from set A to set B. If A and B are the subset of the set of real numbers then f is a real function.

A function f of the form f(x) = for each x ε R is called a modulus function.

If 'x' is a non negative real number, then f(x) = x

If 'x' is a negative number then f(x) = -x

i.e.|4| = 4

|-4| = -(-4) = 4

Domain of the modulus function is the set of all real numbers but the range is the set of all non negative real numbers.

i.e. Domain = R

Range = R+ or [0,∞)

Graph of f(x) =

x

-2

-1

0

1

2

Y= |x|

2

1

0

1

2

9. Greatest integer function

A greatest integer function f(x) = [x] is defined as the value of the greatest integer, less than or equal to x

E.g. [3.2] =3

[5.9] = 5

[-7.2] = -8

Domain =R (Real numbers)

Range = Z (set of integers)

10. Smallest integer function

Smallest integer function f(x) = {x} is defined as the value of smallest integer greater than or equal to x.

E.g. {3.1} = 4

{-7.4}= -7

Domain = R

Range = Z

11. Odd Function

If f (-x) = - f(x) then f is an odd function

E.g. If f(x) = x + 2x3

Then

i.e. f(-2) = -f(2)

f is an odd function

12. Even Function

If f(-x) = f(x) then f is an even function

E.g. If f(x) = 2x2+1

f (1) = 2(1)2+1 = 3

and f(-1) = 2(-1)2+1 = 3

i.e. f(-1) = f(1)

f is an even function

Transcendental functions

A function which cannot be expressed algebraically is called a transcendental function.

Main transcendental functions are explained below.

Trigonometric function

A function defined in the form of a trigonometric ratio is called a trigonometric function

E.g. f(x) = Sin x

g(x) = Cos x

etc

E.g. f(x)= Sin x

Sin x is a periodic function with period 2π.

Domain =R

Range = [-1,1]

Logarithmic function

A function of the form f(x) = log x is called a logarithmic function

E.g f(x) = logex

Domain = R+

Range = R

Main Objectives under this module:

More understanding of the graphs

Transformations of a graph

Vertical shift

Horizontal shift

Stretch and Shrink

Changing directions

Maximum and Minimum value of a function

Application of derivatives in Maxima and Minima

More understanding on the graph of a function.

Transformations of a graph

Vertical shift:

Through the sufficient changes in the function, its corresponding graph can move upward or downward.

Consider the graph of the function f(x) = x2.

f(x) → f(x) + k, where 'k' is a positive real constant.

Then the graph moves 'k' units upward.

f(x) → f(x) - k, where 'k' is a positive real constant.

Then the graph moves 'k' units downward.

Horizontal shift:

f(x) → f(x - k), where 'k' is a positive real constant.

Then the graph moves 'k' units towards right.

f(x) → f(x + k), where 'k' is a positive real constant.

Then the graph moves 'k' units towards left.

Shrink and stretch:

f(x) → k f(x), where k > 1

Then the graph will shrink 'k' times along y axis

f(x) → 1/k * f(x), where k > 1

Then the graph will stretch 'k' units along y axis.

Changing direction:

f(x) → - f(x)

Then the graph will revolve 180° about x- axis.

Maxima and Minima

Here we are discussing the maximum and minimum value of a function. Its conventional application is lying in the area of differential calculus, even though we can think about some of the simple methods irrespective of the application of differentiation for finding the maxima and minima.

Examples:

What is the minimum value of the function x2 + 6x + 12 ?

Solution:

x2 + 6x + 12 = x2 + 6x + 9 + 3

= (x + 3)2 + 3

Now the minimum value for (x +3)2 can be equal to zero, when x = -3.

Hence the minimum value of the function is 0 + 3 = 3.

Alternate method 1: (Formula)

Apply the formula 4ac- b2/4a for finding the minimum value of the given function.

a = 1

b = 6

c = 12

Minimum value = (4*1*12 - 62)/4*1 = 3

Alternate method 2: (differentiation)

Let y = x2 + 6x + 12

y' = dy/dx = 2x + 6

y'' = d2y/dx2 = 2

y'' > 0, hence the function has a minimum value.

Equate y' to zero to get the value of 'x' at which the given function has a minimum value.

2x + 6 = 0

x = -3

y = (-3)2 + 6(-3) + 12 = 3

Hence the minimum value of y = 3

What are the maximum and minimum values of the function y = x2+ 2x+1/x2+ x+1

Solution:

Method 1:

y = x2+ 2x+1/x2+ x + 1

y (x2 x+1) = x2+ 2x+1

(y - 1) + (y - 2) x + (y - 1) = 0

As 'x' is a real number, b2- 4ac ≥ 0

(y - 2)2 - 4 (y -1) 2 ≥ 0

y (3y - 4) ≤ 0

y ≥ 0 and 3y - 4 ≤ 0

y ≥ 0 and y ≤ 4/3

Hence the minimum value of the function is '0' and the maximum value of the function is 4/3.

Method 2:

Application of derivatives:

Quotient rule of differentiation: d[U/V]/dx = (U'V-UV')/V2

Equate the derivative of the function to zero to get the values at which function has the minimum and maximum values.